Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string itosm(long long x) {
  if (x == 0) return "0";
  string res = "";
  while (x > 0) {
    res += ((x % 10) + '0');
    x /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
long long stoim(string str) {
  long long res = 0;
  int p = 0;
  if (str[0] == '-') p++;
  for (int i = p; i < str.length(); i++) {
    res *= 10;
    res += (str[i] - '0');
  }
  return res;
}
const long long infll = 1e18 + 3;
const int inf = 1009000999;
const long double eps = 1e-7;
const int maxn = 1e6 + 1146;
const int baseint = 1000200013;
const long long basell = 1e18 + 3;
const long double PI = acos(-1.0);
const long long mod = 1e9 + 9;
vector<pair<double, int> > q;
double ans[maxn];
pair<double, double> t[maxn];
void solve() {
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    q.push_back({l, 1});
    q.push_back({r, -1});
  }
  int qq;
  cin >> qq;
  for (int i = 0; i < qq; i++) {
    double x, y;
    cin >> x >> y;
    double dx = a - x, dy = y - sy;
    dx *= y / dy;
    t[i].first = dx;
    q.push_back({x + dx, -i - 2});
    dx = b - x;
    dx *= y / dy;
    t[i].second = dx;
    q.push_back({x + dx, i + 2});
  }
  sort(q.begin(), q.end());
  double sum = 0, last = 0;
  for (pair<long double, int> tmp : q) {
    if (tmp.second == 1)
      last = tmp.first;
    else if (tmp.second == -1)
      sum += (tmp.first - last), last = -1;
    else if (tmp.second > 0) {
      ans[tmp.second - 2] += sum;
      if (last > 0) ans[tmp.second - 2] += tmp.first - last;
    } else {
      ans[-tmp.second - 2] -= sum;
      if (last > 0) ans[-tmp.second - 2] -= tmp.first - last;
    }
  }
  cout << fixed << setprecision(7);
  for (int i = 0; i < qq; i++)
    cout << ans[i] / (t[i].second - t[i].first) * (b - a) << "\n";
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}